\(\int \frac {x^8}{(a+c x^4)^3} \, dx\) [681]

Optimal result

Integrand size = 13, antiderivative size = 221 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=-\frac {x^5}{8 c \left (a+c x^4\right )^2}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}-\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}-\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}} \]

[Out]

-1/8*x^5/c/(c*x^4+a)^2-5/32*x/c^2/(c*x^4+a)+5/128*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(3/4)/c^(9/4)*2^(1/2)
+5/128*arctan(1+c^(1/4)*x*2^(1/2)/a^(1/4))/a^(3/4)/c^(9/4)*2^(1/2)-5/256*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)
+x^2*c^(1/2))/a^(3/4)/c^(9/4)*2^(1/2)+5/256*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))/a^(3/4)/c^(9/4)*
2^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {294, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=-\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}-\frac {5 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}-\frac {x^5}{8 c \left (a+c x^4\right )^2} \]

[In]

Int[x^8/(a + c*x^4)^3,x]

[Out]

-1/8*x^5/(c*(a + c*x^4)^2) - (5*x)/(32*c^2*(a + c*x^4)) - (5*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(64*Sqrt
[2]*a^(3/4)*c^(9/4)) + (5*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(64*Sqrt[2]*a^(3/4)*c^(9/4)) - (5*Log[Sqrt[
a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(128*Sqrt[2]*a^(3/4)*c^(9/4)) + (5*Log[Sqrt[a] + Sqrt[2]*a^(1/4
)*c^(1/4)*x + Sqrt[c]*x^2])/(128*Sqrt[2]*a^(3/4)*c^(9/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^5}{8 c \left (a+c x^4\right )^2}+\frac {5 \int \frac {x^4}{\left (a+c x^4\right )^2} \, dx}{8 c} \\ & = -\frac {x^5}{8 c \left (a+c x^4\right )^2}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}+\frac {5 \int \frac {1}{a+c x^4} \, dx}{32 c^2} \\ & = -\frac {x^5}{8 c \left (a+c x^4\right )^2}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}+\frac {5 \int \frac {\sqrt {a}-\sqrt {c} x^2}{a+c x^4} \, dx}{64 \sqrt {a} c^2}+\frac {5 \int \frac {\sqrt {a}+\sqrt {c} x^2}{a+c x^4} \, dx}{64 \sqrt {a} c^2} \\ & = -\frac {x^5}{8 c \left (a+c x^4\right )^2}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}+\frac {5 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{128 \sqrt {a} c^{5/2}}+\frac {5 \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{128 \sqrt {a} c^{5/2}}-\frac {5 \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx}{128 \sqrt {2} a^{3/4} c^{9/4}}-\frac {5 \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx}{128 \sqrt {2} a^{3/4} c^{9/4}} \\ & = -\frac {x^5}{8 c \left (a+c x^4\right )^2}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}-\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}} \\ & = -\frac {x^5}{8 c \left (a+c x^4\right )^2}-\frac {5 x}{32 c^2 \left (a+c x^4\right )}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{3/4} c^{9/4}}-\frac {5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}}+\frac {5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{128 \sqrt {2} a^{3/4} c^{9/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.91 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=\frac {\frac {32 a \sqrt [4]{c} x}{\left (a+c x^4\right )^2}-\frac {72 \sqrt [4]{c} x}{a+c x^4}-\frac {10 \sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {10 \sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{3/4}}-\frac {5 \sqrt {2} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{3/4}}+\frac {5 \sqrt {2} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{3/4}}}{256 c^{9/4}} \]

[In]

Integrate[x^8/(a + c*x^4)^3,x]

[Out]

((32*a*c^(1/4)*x)/(a + c*x^4)^2 - (72*c^(1/4)*x)/(a + c*x^4) - (10*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1
/4)])/a^(3/4) + (10*Sqrt[2]*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(3/4) - (5*Sqrt[2]*Log[Sqrt[a] - Sqrt[2
]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/a^(3/4) + (5*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2
])/a^(3/4))/(256*c^(9/4))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.92 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.24

method	result	size
risch	\(\frac {-\frac {9 x^{5}}{32 c}-\frac {5 a x}{32 c^{2}}}{\left (x^{4} c +a \right )^{2}}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} c +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{128 c^{3}}\)	\(54\)
default	\(\frac {-\frac {9 x^{5}}{32 c}-\frac {5 a x}{32 c^{2}}}{\left (x^{4} c +a \right )^{2}}+\frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 c^{2} a}\)	\(132\)

[In]

int(x^8/(c*x^4+a)^3,x,method=_RETURNVERBOSE)

[Out]

(-9/32/c*x^5-5/32/c^2*a*x)/(c*x^4+a)^2+5/128/c^3*sum(1/_R^3*ln(x-_R),_R=RootOf(_Z^4*c+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.19 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=-\frac {36 \, c x^{5} - 5 \, {\left (c^{4} x^{8} + 2 \, a c^{3} x^{4} + a^{2} c^{2}\right )} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (a c^{2} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} + x\right ) + 5 \, {\left (-i \, c^{4} x^{8} - 2 i \, a c^{3} x^{4} - i \, a^{2} c^{2}\right )} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (i \, a c^{2} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} + x\right ) + 5 \, {\left (i \, c^{4} x^{8} + 2 i \, a c^{3} x^{4} + i \, a^{2} c^{2}\right )} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (-i \, a c^{2} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} + x\right ) + 5 \, {\left (c^{4} x^{8} + 2 \, a c^{3} x^{4} + a^{2} c^{2}\right )} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (-a c^{2} \left (-\frac {1}{a^{3} c^{9}}\right )^{\frac {1}{4}} + x\right ) + 20 \, a x}{128 \, {\left (c^{4} x^{8} + 2 \, a c^{3} x^{4} + a^{2} c^{2}\right )}} \]

[In]

integrate(x^8/(c*x^4+a)^3,x, algorithm="fricas")

[Out]

-1/128*(36*c*x^5 - 5*(c^4*x^8 + 2*a*c^3*x^4 + a^2*c^2)*(-1/(a^3*c^9))^(1/4)*log(a*c^2*(-1/(a^3*c^9))^(1/4) + x
) + 5*(-I*c^4*x^8 - 2*I*a*c^3*x^4 - I*a^2*c^2)*(-1/(a^3*c^9))^(1/4)*log(I*a*c^2*(-1/(a^3*c^9))^(1/4) + x) + 5*
(I*c^4*x^8 + 2*I*a*c^3*x^4 + I*a^2*c^2)*(-1/(a^3*c^9))^(1/4)*log(-I*a*c^2*(-1/(a^3*c^9))^(1/4) + x) + 5*(c^4*x
^8 + 2*a*c^3*x^4 + a^2*c^2)*(-1/(a^3*c^9))^(1/4)*log(-a*c^2*(-1/(a^3*c^9))^(1/4) + x) + 20*a*x)/(c^4*x^8 + 2*a
*c^3*x^4 + a^2*c^2)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.31 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=\frac {- 5 a x - 9 c x^{5}}{32 a^{2} c^{2} + 64 a c^{3} x^{4} + 32 c^{4} x^{8}} + \operatorname {RootSum} {\left (268435456 t^{4} a^{3} c^{9} + 625, \left ( t \mapsto t \log {\left (\frac {128 t a c^{2}}{5} + x \right )} \right )\right )} \]

[In]

integrate(x**8/(c*x**4+a)**3,x)

[Out]

(-5*a*x - 9*c*x**5)/(32*a**2*c**2 + 64*a*c**3*x**4 + 32*c**4*x**8) + RootSum(268435456*_t**4*a**3*c**9 + 625,
Lambda(_t, _t*log(128*_t*a*c**2/5 + x)))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.96 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=-\frac {9 \, c x^{5} + 5 \, a x}{32 \, {\left (c^{4} x^{8} + 2 \, a c^{3} x^{4} + a^{2} c^{2}\right )}} + \frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{256 \, c^{2}} \]

[In]

integrate(x^8/(c*x^4+a)^3,x, algorithm="maxima")

[Out]

-1/32*(9*c*x^5 + 5*a*x)/(c^4*x^8 + 2*a*c^3*x^4 + a^2*c^2) + 5/256*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x +
sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))) + 2*sqrt(2)*arctan(1/2*sqrt(2
)*(2*sqrt(c)*x - sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))) + sqrt(2)*log
(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(1/4)) - sqrt(2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1
/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(1/4)))/c^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.92 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=\frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{128 \, a c^{3}} + \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{128 \, a c^{3}} + \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{256 \, a c^{3}} - \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{256 \, a c^{3}} - \frac {9 \, c x^{5} + 5 \, a x}{32 \, {\left (c x^{4} + a\right )}^{2} c^{2}} \]

[In]

integrate(x^8/(c*x^4+a)^3,x, algorithm="giac")

[Out]

5/128*sqrt(2)*(a*c^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^3) + 5/128*sqrt(2
)*(a*c^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^3) + 5/256*sqrt(2)*(a*c^3)^(1
/4)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a*c^3) - 5/256*sqrt(2)*(a*c^3)^(1/4)*log(x^2 - sqrt(2)*x*(a/
c)^(1/4) + sqrt(a/c))/(a*c^3) - 1/32*(9*c*x^5 + 5*a*x)/((c*x^4 + a)^2*c^2)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.37 \[ \int \frac {x^8}{\left (a+c x^4\right )^3} \, dx=-\frac {\frac {9\,x^5}{32\,c}+\frac {5\,a\,x}{32\,c^2}}{a^2+2\,a\,c\,x^4+c^2\,x^8}-\frac {5\,\mathrm {atan}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{64\,{\left (-a\right )}^{3/4}\,c^{9/4}}-\frac {5\,\mathrm {atanh}\left (\frac {c^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )}{64\,{\left (-a\right )}^{3/4}\,c^{9/4}} \]

[In]

int(x^8/(a + c*x^4)^3,x)

[Out]

- ((9*x^5)/(32*c) + (5*a*x)/(32*c^2))/(a^2 + c^2*x^8 + 2*a*c*x^4) - (5*atan((c^(1/4)*x)/(-a)^(1/4)))/(64*(-a)^
(3/4)*c^(9/4)) - (5*atanh((c^(1/4)*x)/(-a)^(1/4)))/(64*(-a)^(3/4)*c^(9/4))